Why do a lot of #defines in the Linux kernel use do { ... } while(0)?

There are a couple of reasons:

• Empty statements give a warning from the compiler so this is why you see #define FOO do { } while(0).
• It gives you a basic block in which to declare local variables.
• It allows you to use more complex macros in conditional code. Imagine a macro of several lines of code like:

#define FOO(x) \
printf("arg is %s\n", x); \
do_something_useful(x);
Now imagine using it like:

if (blah == 2)
FOO(blah);

This interprets to:

if (blah == 2)
printf("arg is %s\n", blah);
do_something_useful(blah);;

As you can see, the if then only encompasses the printf(), and the do_something_useful() call is unconditional (not within the scope of the if), like you wanted it. So, by using a block like do { ... } while(0), you would get this:

if (blah == 2)
do {
printf("arg is %s\n", blah);
do_something_useful(blah);
} while (0);
Which is exactly what you want.

If you want a block statement so you can have several lines of code and declare local variables. But then the natural thing would be to just use for example:

#define exch(x,y) { int tmp; tmp=x; x=y; y=tmp; }
However that wouldn't work in some cases. The following code is meant to be an if-statement with two branches:

if (x > y)
exch(x,y); // Branch 1
else
do_something(); // Branch 2

But it would be interpreted as an if-statement with only one branch:

if (x > y) { // Single-branch if-statement!!!
int tmp; // The one and only branch consists
tmp = x; // of the block.
x = y;
y = tmp;
}
; // empty statement
else // ERROR!!! "parse error before else"
do_something();

The problem is the semi-colon (;) coming directly after the block. The solution for this is to sandwich the block between do and while (0). Then we have a single statement with the capabilities of a block, but not considered as being a block statement by the compiler. Our if-statement now becomes:

if (x > y)
do {
int tmp;
tmp = x;
x = y;
y = tmp;
} while(0);
else
do_something();